Let $X$ be a projective variety over $\mathbb{C}$ and let $T=(\mathbb{C^*})^k$ act on it.
Is it true that there is a fixed point of this action on every irreducible component of $X$ just because $X$ is projective?
If not are there some mild condition such that I can deduce that?
I believe the answer is yes. Here is an argumen
Let $\mathcal{O}$ be any orbit of your group action. Then the closure $\overline{\mathcal{O}}$ is a union of $\mathcal{O}$ and lower dimensional orbits. This is because $\mathcal{O}$ is a constructible subset of $X$ as it is the image of the morphism $T \to X$, $t \mapsto t\cdot x$ for any fixed $x \in \mathcal{O}$. Then it contains some open subset $U$ of its closure since it is a finite union of locally closed subsets. But $T$ acts transitively by isomorphisms so
$$ \mathcal{O} = \bigcup_{t\in T} t(U) $$
where $t(U)$ are open in the closure. Then $\mathcal{O}$ itself is open in its closure so $\overline{\mathcal{O}} \setminus \mathcal{O}$ is closed and $T$-invariant so it must be a union of lower dimensional orbits.
In particular, any orbit of minimum dimension is closed so closed orbits must exist. Let $\mathcal{O}$ be one such orbit. Since $X$ is projective, then $\mathcal{O}$ is also projective but it is also isomorphic to the image of the map $T \to X$ described above. In fact for $x \in \mathcal{O}$, $\mathcal{O} \cong T/T_x$ where $T_x$ is the stabilizer of $x$. So $T/T_x$ is projective but a quotient of algebraic toric is always a torus of dimension $\dim T - \dim T_x$. The only projective algebraic torus is the trivial one of dimension $0$ and so $\dim T = \dim T_x$, $T = T_x$, and the orbit $\mathcal{O}$ is exactly just the point $x$. That is, $x$ is a torus fixed point.